College of Engineering and Computer Science
Mechanical Engineering Department
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
1 Complete the following table for H
2
O
Row
1
2
3
4
T( o C)
50
250
110
P(kPa) v(m 3 /kg) Phase description
200
4.16
Saturated vapor
400
600
Use the Tables A-4 through A-8 for water starting on page 914. For the first row data, we first try to see if the given specific volume of 4.16 m 3 /kg lies in the liquid, gas, or mixed region. We do this by looking at the values of v f
and v g
at the given temperature of 50 o C. On page 914, we find that v f
(50 o C) = 0.001012 m 3 /kg and v g
(50 o C) = 12.026 m 3 /kg. Since the given volume of 4.16 m 3 /kg lies between the saturated liquid volume, v f
, and the saturated vapor volume, v g
, at the given temperature of 50 o C, we conclude that the state is in the mixed region. Since we are in the mixed region, the pressure is the saturation pressure at the given temperature of 50 o C. From the same location in Table A-4 we find P sat
(50 o C) = 12.352 kPa. The quality at this point can be found as follows. x
 v v
 v f g
 v f
4 .
16 m
3 kg

12 .
026 m
3 kg

0 .
001012 m
3 kg

0 .
001012 m
3 kg

0 .
346
The point in row two is defined as saturated vapor. This means that the temperature is the saturation pressure at 200 kPa and the volume is the volume of the saturated vapor at this pressure. From Table A-5 on page 916 we find T sat
(200 kPa) = 120.21
o C and v g
(200 kPa) =
0.88578 m 3 /kg.
The data in row three are found in the superheat table, A-6, on page 918; the volume at this point is 0.59520 m 3 /kg. Note that we the pressures in this table are in MPa so we have to recognize that 400 kPa = 0.4 MPa.
In searching for the state in row four, we find that the temperature of 110 o C is less than the saturation temperature at 600 kPa, which is found in Table A-5 on page 916 to be 158.83
o C.
This can also be found if we guessed that this was a gas and tired to locate this pressure and temperature in the superheat region: Table A-6 on page 918. We see that that the minimum temperature in the section of this table for 600 kPa (stated as 0.6 MPa) is 200 o C. To see if this is a liquid we can find the saturation temperature at the top of the table section, next to the 0.6 MPa value. Because the given temperature of 110 o C is less than the saturation temperature of
158.83
o C, the state is a compressed liquid. The simplest approach is to approximate the specific volume at this state as the volume of the saturated liquid at the given temperature of 110 o C.
This volume is found to be 0.001052 m 3 /kg (Table A-4, page 914).
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062

Unit one homework solutions ME 370, L. S. Caretto, Fall 2010 Page 2
The alternative approach would be a double interpolation in the compressed liquid table on page
924. Because this table does not have a temperature of 110 o C, we have to interpolate data between the table temperatures of 100 o C and 120 o C. Because the minimum pressure in this table is 5 MPa, we have to interpolate between the saturation pressure and the pressure of 5
MPa to get the values at the desired pressure of 600 kPa.
First interpolate to find the value at 100 o C and 600 kPa.
 v

100 o
C , 600
0 .
001052 kg kPa
  f
100 o
C
 
100 o
C
5000 m
3

0 .
0010410 kg
5000 m
3 kPa


101 .
42
, 5000
0 .
001052 kg kPa kPa m
3 kPa

P sat
 
 f
100 o
100
C
 o
C


600 kPa

101 .
42

600 kPa
 kPa


P sat

100 o
C

0 .
0010509 kg m
3

Next interpolate to find the value at 120 o C and 600 kPa.
 v

120 o
C , 600
0 .
001060 kg m
3 kPa
  f
120 o
C
 
120 o
C
5000
, 5000 kPa kPa

P sat
 
 f
120 o
120
C
 o
C

0 .
0010576 m
3

0 .
001060 m
3
 kg
5000 kPa

198 .
67 kg kPa

600 kPa

198 .
67

600 kPa
 kPa


P sat

120 o
C

0 .
0010598 kg m
3

Finally we can interpolate between the 100 o C and 120 o C at the pressure of 600 kPa. v

110 o

C , 600 kPa
 
100
0 .
0010509 kg m
3

0 .
o
C , 600
0010598 kg kPa
 
120
120 m o
3
C


0 .
o
C , 600 kPa
 
100 o
0010509
120 m
3 o
100 o
C kg

C

110
100 o
C o

C
C ,
100
600 o
C

 kPa

110 o
C
0 .
0010553 kg

100 o
C
 m
3
We see that this answer is close to the value of 0.001052 m 3 /kg for the saturated liquid volume at the given temperature of 110 0 C.
The answers for this row and the previous rows are shown in the table below.
Row
1
2
3
4
T( o C)
50
120.23
250
110
P(kPa)
12.352
200
400 v(m 3 /kg) Phase description
4.16 Mixed region with x = 0.346
0.88578 Saturated vapor
0.59520 Superheated vapor
600 0.001052 Compressed liquid

Unit one homework solutions ME 370, L. S. Caretto, Fall 2010 Page 3
2 Complete the following table for refrigerant 134a:
Row
1
2
3
4
5
T( o F) P(psia) h(Btu/lb m
) x
15
80 78
0.6
10
110
70
180 129.467
1.0
Phase description
Use the Tables A-11E through A-13E for refrigerant 134a starting on page 976. For the first row data, we find that the specific enthalpy of 78 Btu/lb m
lies between the saturated liquid and saturated vapor enthalpies at the given pressure of 80 psia. (On page 9777, Table A12E , we find that h f
(80 psia) = 33.394 Btu/lb m
and h g
(80 psia) = 112.20 Btu/lb m
; the difference between these two, h fg
(80 psia) = 78.804 Btu/lb m
is given in the table.) Thus, the state is in the mixed region and the temperature is the saturation temperature at the given pressure of 80 psia. From the same location in Table A12E we find T sat
(80 psia) = 65.89
o F. The quality at this point can be found as follows. x
 h
 h fg h f 
78
Btu lb m

33 .
394
Btu
78 .
804 lb m
Btu lb m 
0 .
566
The point in row two has a quality of 0.6. Thus, it must be in the mixed region. We use Table A-
11E on page 976 to get the saturation data. The pressure is the saturation pressure at the given temperature, P sat
(15 o F) = 29.759 psia, and we compute the enthalpy from the given quality and the tabulated values of enthalpy of the saturated liquid, h f
(15 o F) = 16.889 Btu/lb m
, and the enthalpy change between saturated liquid and saturated vapor, h fg
(15 o F) = 88.377 Btu/lb m
. h
 h f
 xh fg

16 .
889
Btu lb m

0 .
6 88 .
377
Btu lb m

69 .
915
Btu lb m
If we look for the data in row three (T = 10 o F, P = 70 psia) in the superheat table on page 978, we find that the saturation temperature at 70 psia is 58.30
o F. Since the given temperature is less than this we know that we have a compressed liquid. We can approximate the compressed liquid properties as those of the saturated liquid state at the given temperature of 10 o F. From Table
A-11E on page 976, we find that h f
(10 o F) = 15.318 Btu/lb m
.
The data in row four are found in the superheat table on page 979. We look for an enthalpy of
129.46 Btu/lb m
in the small table for a pressure of 180 psia and find that this occurs at a temperature of 160 o F. (Normally we would not be so lucky and would have to interpolate.)
The data in the last row specify a quality of one. This means that we have a saturated vapor. We use the saturation table on page 976 to find P = P sat
(110 o F) = 161.16 kPa and h g
(110 o F) =
117.23 Btu/lb m
. The results for all rows are summarized in the table below.
Row
1
2
3
4
5
T( o F) P(psia) h(Btu/lb m
) x
65.89
15
80
29.759
78
69.915
0.566
0.6
10
160
110
70
180
161.16
15.318
129.46
117.23
N/A
N/A
1.0
Phase description
Mixed region
Mixed region
Compressed liquid
Superheated vapor
Saturated vapor

Unit one homework solutions ME 370, L. S. Caretto, Fall 2010 Page 4
3 A piston-cylinder device initially contains 50L of liquid water at 25 o C and 300 kPa. Heat is added to the water at constant pressure until the entire liquid is vaporized. (a) What is the mass of the water? (b) What is the final temperature? (c) Determine the total enthalpy change. (d) Show the process on a T-v diagram with respect to saturation lines.
We find that the initial state of 25 o C and
300 kPa is a compressed liquid, because the temperature of 25 o C is less than the saturation temperature at 300 kPa. (See
Table A-5 on page 916 that gives T sat
(300
300 kPa = 133.52
o C. Alternatively we could find that the pressure of 300 kPa is greater than the saturation pressure at
25 o C, which is found as 3.1598 kPa in
Table A-4 on page 915.) We approximate this compressed liquid state as the saturated liquid at the given temperature of 25 o C, from Table A-4. This gives an initial volume, v
1
≈ v f
0.001003 m 3
(25 o C) =
/kg and an initial enthalpy, h
1
≈ h f
(25 o C) = 104.89 kJ/kg.
1000
100
10
1
initial point final point
Since we know the total volume, V = 50L =
0.05 m 3 and the initial specific volume, v
1
=
0.001003 m 3 /kg, we can find the mass in the system as the ratio of these two quantities.
0.1
0.0001
0.01
1 100
Specific volume (m3/kg)
10000 m

V
1 v
1
0 .
05 m
3

0 .
001003 m
3 kg m = 49.85 kg
We are told that the entire liquid is vaporized at the final state. If we assume that there is no superheat, the final state must be that of saturated vapor. Since we are also told that the process occurs at constant pressure, the final pressure, P
2
= P
1
= 300 kPa. The final temperature is the saturation temperature at 300 kPa from Table A-5; i.e
., T
2
= T sat
(300 kPa) = 133.55C
.
The total enthalpy change is the difference between the initial and final enthalpy, multiplied by the system mass. We found the initial enthalpy above; we know that the final state is a saturated vapor at 300 kPa; thus h
2
= h g
(300 kPa) = 2724.9 kJ/kg, and the total enthalpy change is given by the following equation.

H = m(h
2
– h
1
) = (49.85 kg)(2724.9 kJ/kg – 104.89 kJ/kg = 1.306x10
5 kJ/kg
The T-v diagram with the process indicated is shown above. On this diagram the process is a solid line (blue if you view this in color) and the saturation curve is the dotted line (purple in color).
The temperature of the compressed liquid increases, at constant pressure, until the saturation temperature at the constant pressure of 300 kPa (133.55
o C) is reached. This initial part of the process is so close to the saturation line that we cannot distinguish the two on the plot. Once the saturation temperature is reached, the additional heat vaporizes the liquid water. This part of the

Unit one homework solutions ME 370, L. S. Caretto, Fall 2010 Page 5 process takes place at constant temperature. (In the mixed region a constant pressure process is a constant temperature pressure process and vice versa .)
4 The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25 o C, the pressure gage reads 210 kPa. If the volume of the tire is
0.025 m 3 , determine the pressure rise in the tire when the temperature in the tire rises to
50 o C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa.
For this problem we can use the ideal gas equation. The gas constant for air is found from Table
A-1 on page 908 to be 0.2870 kJ/kg ·. At the initial state with an atmospheric pressure of 100 kPa and a gage pressure of 210 kPa, the absolute pressure is 210 kPa + 100 kPa = 310 kPa. The absolute temperature initially is 25 + 273.15 = 298.15 K. The absolute temperature at the final state is 50 + 273.15 = 323.15 K.
In the first part of this problem, the mass is constant. We are not told anything about how the tire expands as the temperature increases. Let’s make the simple assumption that the volume remains constant.
According to the ideal gas equation, PV = mRT. If we compare two states for which mass and volume are the same we have the following result.
V mR
 const

T
1
P
1

T
2
P
2

P
2

P
1
T
2
T
1
Substituting the given data into the final equation gives the pressure at the temperature of 50 o C.
P
2

P
1
T
2
T
1

( 310 kPa )
323 .
15
298 .
15
K
K

336 kPa
The pressure rise is simply the difference P
2
– P
1
= (336 – 310) kPa = 26 kPa .
We contemplate a third state, after we bleed off the air. In this final state the pressure equals the initial pressure (P
3
= P
1
= 310 kPa), while the final temperature remains at 50 o C. (T
3
= T
2
=
50 o C.) The mass difference between these two states is computed from the ideal gas equation as follows.
 m
 m
1
 m
3

P
1
V
RT
1

P
3
V
RT
3

P V
1
R


1
T
1

1
T
3



( 310 kPa )( 0 .
025 m
3
)
0 .
2870 kJ


1
298 .
15 K kg K

1
323 .
15 K


 m = 0.0070 kg

Unit one homework solutions ME 370, L. S. Caretto, Fall 2010 Page 6
5 A rigid tank contains 20 lb m
of air at 20 psia and 70 o F. More air is added to the tank until the pressure and temperature rise to 35 psia and 90 o F, respectively. Determine the amount of air added to the tank.
We can solve this problem by assuming that air is an ideal gas. The gas constant for air is given as 0.06855 Btu/(lb m
•R) = 0.3704 psia•ft 3 /(lb m
•R) in Table A-1E on page 958 of the text. The latter value and units are used here because they are consistent with the units specified for other variables.
If we denote the initial state as state 1 and the final state as state 2, we can apply the ideal-gas equation of state to both the initial and final state as shown below. m
1

P
1
V
1
RT
1 m
2

P
2
V
2
RT
2
In this problem we are given m
1
= 20 lb m
, P
1
= 20 psia, T
1
= 70 o F = 529.67 R, P
2
= 35 psia, T
2
=
90 o F = 549.67 R. We are also told that the contained is rigid so that we can assume that V
2
= V
1
.
Use the symbol, V = V
1
= V
2
, to denote this constant volume. If we solve the ideal-gas equation at state 1 for this constant volume, V, and substitute the result into the equation at state 2 we get the following equation for the mass added.
 m
 m
2
 m
1

P
2
V
2
RT
2
 m
1

P
2 m
1
RT
P
1
RT
2
1
 m
1
 m
1
P
2
T
P
1
T
2
1  m
1
 m
1


P
2
T
1
P
1
T
2

1


Substituting the known values of P
1
, T
1
, P
2
, T
2
, and m
1
into this equation gives the final result.
 m
 m
1


P
2
P
1
T
1
T
2

1



( 20 lb m
)


( 35
( 20 psia )( 529 .
67 R ) psia )( 549 .
67 R )

1

 = 13.73 lb m

Unit one homework solutions ME 370, L. S. Caretto, Fall 2010 Page 7
6 Determine the specific volume of superheated water vapor at 10 MPa and 400 o C using (a) the ideal gas equations, (b) the generalized compressibility charts and (c) the steam tables. Also, determine the error involved in the first two cases.
We can solve part (c) first. The steam-table specific volume is found from Table A-6 on page 908 for the given pressure and temperature. v steam tables
= 0.026436 m 3 /kg
To use the ideal-gas equation of state, we first obtain the gas constant for H
2
O from table A-1 on page 88 4: R = 0.4615 kJ/(kg•K) for water. Applying this to the pressure of 10 MPa = 10,000 kPa and a temperature of 400 o C = 673.15 K gives the following specific volume. v ideal gas

RT
P
0 .
4615 kg

K kJ
( 673 .
15 K )

1 kJ
( 10 , 000 kPa ) kPa
 m
3
= 0.03106 m 3 /kg
The error in the ideal gas calculation is found by comparing the result to the specific volume obtained from the steam tables.
Error

0 .
03106 m
3 kg

0 .
026416 m
3 kg
0 .
026416 m
3 kg
= 17.6%
In order to use the generalized compressibility chart, we first have to compute the reduced pressure and temperature. These are simply the ratio of the given pressure and temperature to the critical pressure and temperature, respectively. We find the critical values for water from
Table A-1 on page 908: T c
= 647.1 K and P c
= 22.06 MPa. The error in the ideal gas calculation is found by comparing the result to the specific volume obtained from the steam tables.
P r

P
P c

10 MPa
22 .
06 MPa

0 .
453 T r

T
T c

673 .
15 K
647 .
1 K

1 .
04
With these values of reduced pressure and temperature, we can find the compressibility factor from the chart in Figure A-15 on page 932. This gives Z = 0.84. With this value of Z, we can find the desired specific volume as follows. v
Z chart

ZRT
P

Zv ideal gas

( 0 .
84 ) 0 .
03106 m
3 kg
= 0.026 m 3 /kg
As with the ideal-gas calculation, the error is found by comparing the result to the specific volume obtained from the steam tables.
Error

0 .
026 m
3 kg

0 .
026416 m
3 kg
0 .
026416 m
3 kg
= 1.2%